thermo

1. Lecture # 5
PROPERTY TABLES(cont.)
Saturated Liquid–Vapor Mixture
During a vaporization process, a substance exists as part
liquid and part vapor. That is, it is a mixture of saturated liquid
and saturated vapor (Fig. 1. 4).
FIGURE 1- 4
The relative amounts of liquid and vapor phases in a
saturated mixture are specified by the quality

2. Saturated Liquid–Vapor Mixture
During a vaporization process, a substance exists as part
liquid and part vapor. That is, it is a mixture of saturated liquid
and saturated vapor (Fig. 1. 4).
FIGURE 1- 4
The relative amounts of liquid and vapor phases in a
saturated mixture are specified by the quality

3. •To analyze this mixture properly, we need to know the
proportions of the liquid and vapor phases in the mixture.
This is done by defining a new property called the quality x
as the ratio of the mass of vapor to the total mass of the
mixture.
Where:
fg
f
avg
v
v
v
x



4. The analysis given above can be repeated for internal
energy and enthalpy with the following results:
where y is v, u, or h. The subscript “avg” (for “average”) is
usually dropped for simplicity. The values of the average
properties of the mixtures are always between the values
of the saturated liquid and the saturated vapor properties.
That is,

9. EXAMPLE 1-4
An 80-L vessel contains 4 kg of refrigerant-134a at a pressure
of 160 kPa. Determine (a) the temperature, (b) the quality, (c)
the enthalpy of the refrigerant, and (d) the volume occupied
by the vapor phase.
Solution A vessel is filled with refrigerant-134a. Some
properties of the refrigerant are to be determined.
we can determine the specific volume as follows:
At 160 kPa, we also read
vf = 0.000687 m3/kg (Table A- 12)
vg = 0.1031 m3/kg

10. Obviously, vf > v > vg, and, the refrigerant is in the saturated
mixture region. Thus, the temperature must be the saturation
temperature at the specified pressure:
T =Tsat @ 160 kPa = -18.49 ͦC
(b) Quality can be determined from
1886
.
0
000687
.
0
1031
.
0
000687
.
0
02
.
0






fg
f
v
v
v
x
(c) At 160 kPa, we also read from Table A–12 that
hf = 19.18 kJ/kg and hfg = 160.23 kJ/kg. Then,
h = hf +xhfg
= 19.18 (kJ/kg) + 0.189 (160.23 kJ/kg)
= 49.46 kJ/kg

11. (d) The mass of the vapor is
mg = xmt = (0.1886) (4 kg) = 0.7544 kg
and the volume occupied by the vapor phase is
Vg = mg vg = (0.7544 kg) (0.1031 m3/kg) = 0.0778 m3
The rest of the volume is occupied by the liquid.
Property tables are also available for saturated solid–
vapor mixtures.
Properties of saturated ice–water vapor mixtures, for
example, are listed in Table A–8. Saturated solid–vapor
mixtures can be handled just as saturated liquid–vapor
mixtures

12. Superheated Vapor:
•In the region to the right of the saturated vapor line and at
temperatures above the critical point temperature, a
substance exists as superheated vapor.
•In these tables, the properties are listed against temperature
for selected pressures starting with the saturated vapor data.
•The saturation temperature is given in parentheses following
the pressure value.

13. •Higher temperatures (T ˃ Tsat at a given P)
•Higher specific volumes (v ˃ vg at a given P or T)
•Higher internal energies (u ˃ ug at a given P or T)
•Higher enthalpies (h ˃ hg at a given P or T)
•Compared to saturated vapor, superheated vapor
is characterized by Lower pressures (P˂ Psat at a
given T )

14. EXAMPLE
Determine the internal energy of water at 20
psia and 400°F.
Solution The internal energy of water at a specified
state is to be determined.
Analysis At 20 psia, the saturation temperature is
227.96°F. Since T ˃ Tsat, the water is in the
superheated vapor region. Then the internal energy
at the given temperature and pressure is
determined from the superheated vapor table
(Table A–6E) to be u=1145.1 Btu/lbm

15. Determine the temperature of water at a
state of P 0.5 MPa and h =2890 kJ/kg.
Solution:
The temperature of water at a specified state
is to be determined.
Example:

16. Analysis At 0.5 MPa, the enthalpy of saturated
water vapor is hg = 2748.7 kJ/kg. Since h ˃ hg, as
shown in the figure, we again have superheated
vapor. Under 0.5 MPa in Table A–6 we read
Obviously, the temperature is between 200 and 250°C.
By linear interpolation it is determined to be
T = 216.3°C

17. Compressed Liquid
•Compressed liquid tables are not as commonly available, and
Table A–7 is the only compressed liquid table in this text.
•The format of Table A–7 is very much like the format of the
superheated vapor tables.
•In the absence of compressed liquid data, a general
approximation is to treat compressed liquid as saturated
liquid at the given temperature. This is because the
compressed liquid properties depend on temperature much
more strongly than they do on pressure.

18. EXAMPLE
Determine the internal energy of compressed liquid water at
80°C and 5 MPa, using
(a) data from the compressed liquid table and (b) saturated
liquid data. What is the error involved in the second case?
Solution The exact and approximate values of the internal
energy of liquid water are to be determined
Analysis At 80°C, the saturation pressure of water is 47.416 kPa,
and since 5 MPa >| Psat, we obviously have compressed liquid, as
shown in the Figure.
(a) From the compressed liquid table (Table A–7)
At P = 5 MPa, T = 80ͦC u = 333.72 kJ/kg
(b) From the saturation table (Table A–4), we read
u ≈ uf@80ͦ C = 334.86 kJ/kg
The error involved is

19. %
34
.
0
100
82
.
333
72
.
333
86
.
334



which is less than 1 percent